题目
设 D: 0 leq r leq 1, 0 leq theta leq (pi)/(2)，根据二重积分的几何意义，iint_(D) sqrt(1-r^2) r dr dtheta = ( ). A. (1)/(6) piB. (1)/(3) piC. (1)/(8) piD. (1)/(4) pi
设 $D: 0 \leq r \leq 1, 0 \leq \theta \leq \frac{\pi}{2}$，根据二重积分的几何意义，$\iint_{D} \sqrt{1-r^2} r dr d\theta = (\ )$.
- A. $\frac{1}{6} \pi$
- B. $\frac{1}{3} \pi$
- C. $\frac{1}{8} \pi$
- D. $\frac{1}{4} \pi$
题目解答
答案
将二重积分转换为累次积分：
\[
\iint\limits_{D} \sqrt{1-r^2} r \, dr \, d\theta = \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} \sqrt{1-r^2} r \, dr \, d\theta
\]
令 $u = 1 - r^2$，则 $du = -2r \, dr$，积分变为：
\[
\int_{0}^{1} \sqrt{1-r^2} r \, dr = \frac{1}{2} \int_{1}^{0} \sqrt{u} \, (-du) = \frac{1}{2} \int_{0}^{1} u^{1/2} \, du = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} \bigg|_0^1 = \frac{1}{3}
\]
代入外积分：
\[
\int_{0}^{\frac{\pi}{2}} \frac{1}{3} \, d\theta = \frac{1}{3} \theta \bigg|_0^{\frac{\pi}{2}} = \frac{\pi}{6}
\]
答案：$\boxed{A}$
解析
步骤 1：转换为累次积分
将二重积分转换为累次积分，即 \[ \iint\limits_{D} \sqrt{1-r^2} r \, dr \, d\theta = \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} \sqrt{1-r^2} r \, dr \, d\theta \]
步骤 2：内积分的变量替换
令 $u = 1 - r^2$，则 $du = -2r \, dr$，积分变为： \[ \int_{0}^{1} \sqrt{1-r^2} r \, dr = \frac{1}{2} \int_{1}^{0} \sqrt{u} \, (-du) = \frac{1}{2} \int_{0}^{1} u^{1/2} \, du \]
步骤 3：计算内积分
计算内积分： \[ \frac{1}{2} \int_{0}^{1} u^{1/2} \, du = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} \bigg|_0^1 = \frac{1}{3} \]
步骤 4：计算外积分
代入外积分： \[ \int_{0}^{\frac{\pi}{2}} \frac{1}{3} \, d\theta = \frac{1}{3} \theta \bigg|_0^{\frac{\pi}{2}} = \frac{\pi}{6} \]
将二重积分转换为累次积分，即 \[ \iint\limits_{D} \sqrt{1-r^2} r \, dr \, d\theta = \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} \sqrt{1-r^2} r \, dr \, d\theta \]
步骤 2：内积分的变量替换
令 $u = 1 - r^2$，则 $du = -2r \, dr$，积分变为： \[ \int_{0}^{1} \sqrt{1-r^2} r \, dr = \frac{1}{2} \int_{1}^{0} \sqrt{u} \, (-du) = \frac{1}{2} \int_{0}^{1} u^{1/2} \, du \]
步骤 3：计算内积分
计算内积分： \[ \frac{1}{2} \int_{0}^{1} u^{1/2} \, du = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} \bigg|_0^1 = \frac{1}{3} \]
步骤 4：计算外积分
代入外积分： \[ \int_{0}^{\frac{\pi}{2}} \frac{1}{3} \, d\theta = \frac{1}{3} \theta \bigg|_0^{\frac{\pi}{2}} = \frac{\pi}{6} \]